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I want to find the amplification, $$U_{out}/U_{in}$$ of the pictured circuit. My electronics knowledge is very basic, but I tried to solve this using the ground node method, with the bottom of the circuit being grounded. I noticed that

$$V_y=U_{out}$$ $$V_1=U_{in}-V_x$$

Then I write equations for the node voltages (I switched resistances to conductances for clarity):

$$V_x(G_E +g_{CE}+g_{be})-V_y (g_{CE}) = gV_1 = gU_{in}-gV_x$$ $$V_y (g_{CE}+G_C)-V_x (g_{CE})=-gV_1=gV_x-gU_{in}$$

Adding the two equations I got:

$$V_x (G_E+g_{be})=-V_y (G_C) \implies V_x=-\frac{G_C}{G_E +g_{be}}V_y$$

Now from the first equation, after substituting:

$$V_y \frac{G_C (G_E+g_{CE}+g_{be})}{g_{be}+G_E} =-gU_{in}$$

$$\frac{V_y}{U_{in}}=\frac{U_{out}}{U_{in}}=-g\frac{ g_{be}+G_E }{G_C (G_E+g_{CE}+g_{be})}$$

Is this solution correct?

I also want to find the input resistance, but here I'm completely out of ideas.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Big applause for going this route but if thisn't homework AND it's the gain of a BJT you are looking for then there are a lot easier ways to come to the answer. \$\endgroup\$ – Andy aka Nov 15 '13 at 21:51
  • \$\begingroup\$ It's a past problem from an exam. I'm 99% sure the problem is supposed to be solved this way. Are my calculations correct then? \$\endgroup\$ – Spine Feast Nov 15 '13 at 22:42
  • \$\begingroup\$ Sorry, I glassed over when you went to admittance but the gain of a common emitter BJT is (ignoring slope of collector current (r_CE)) Rc/Re - that's what I remember. Sorry to be of not much help. \$\endgroup\$ – Andy aka Nov 15 '13 at 23:18
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This is (for the time being) more a comment than an answer.

The first KCL equation should be:

$$(V_x - U_{in})g_{be} + (V_x - V_y)g_{ce} + V_xG_e = (U_{in} - V_x)g$$

which simplifies to:

$$V_x(g_{be} + g_{ce} + G_e + g) -V_yg_{ce} = U_{in}(g_{be} + g) $$

For input resistance, find the current through \$r_{be}\$ which will be in terms of \$U_{in}\$.

The input resistance is then easily found to be:

$$r_{in} = \dfrac{U_{in}}{i_{r_{be}}}$$

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  • \$\begingroup\$ Ok, I wrote down the two equations, but the answer I'm getting is absolutely hideous. Is it meant to be this bad or did I make a mistake in my calculations? \$\endgroup\$ – Spine Feast Nov 16 '13 at 20:08
  • \$\begingroup\$ @DepeHb, which answer; the gain or the input resistance? \$\endgroup\$ – Alfred Centauri Nov 16 '13 at 21:18
  • \$\begingroup\$ The gain. As for the input resistance, I need to have the gain in order to express the current in terms of the input voltage. \$\endgroup\$ – Spine Feast Nov 16 '13 at 21:30
  • \$\begingroup\$ @DepeHb, the actual exact formula for the gain is quite involved however, a very good approximation of the gain is quite simple if you consider the most significant contributions. I'll follow-up later with a calculation of the exact and approximate gains. By the way, you don't need the gain to find the input resistance. \$\endgroup\$ – Alfred Centauri Nov 16 '13 at 22:00

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